The Involute of a Circle

A strong understanding of mathematics is essential in physics. The purpose of these posts are to familiarize the reader with examples of how mathematics, particularly calculus, can be used to solve problems.

This post will discuss an interesting type of pattern called an involute. Using integration I will demonstrate in detail how to calculate its area and its applications to real world problems.

Consider the following scenario about a grazing cow on a farm.

You have a round silo on your farm with a radius of r. Attached on west side of the silo theres a rope of length πr. Tied to the rope, is a cow. What is the total grazing area of the cow?

The pattern outlined by the maxium distance the rope can reach is called an involute. The following is an image of the involute of a circle.

Involute of A Circle

In this example the involutes occur when the rope is past north and south, in the eastward direction. When the rope is at a tangent to the silo towards the north or south, travelling in the western direction, the area is simply just a half circle with radius πr. Therefore the total grazing area is the sum of the area of the two involutes and the half circle.

     \[ A_{\text{\tiny $total$}} = A_{ \text{ \tiny $half circle$ } } + 2 A_{ \text{ \tiny $involute$}} \]

In order to solve this problem we need to express the function as a set of parametric equations. The following is a more detailed representation of the involute.

graph of an involute of a circle

The point T can be described using the coordinates (r cos Θ, r sin Θ ). It can be shown that the line segment TP has a length of rΘ since it was unwound from the arc TA. In order to determine the set of parametric equations we need to first determine the angle PTQ.

     \begin{align*}   \angle PTQ & = \angle PTR - \angle QRT \\              & =  \frac{1}{2} \pi - \theta \end{align*}

With this, it can be shown that the set of parametric equations representing the involute of a circle are as follows.

     \[ \in 0 \le \theta \le \pi \\ \] \begin{align*}   x & = r cos \theta + r \theta cos \left( \frac{1}{2} \pi - \theta \right) \\     & = r (  cos \theta + \theta sin \theta ) \\   y & = r sin \theta - r \theta sin \left( \frac{1}{2} \pi - \theta \right) \\     & = r (  sin \theta - \theta cos \theta ) \\ \end{align*}

If we differentiate the two parametric equations we obtain the following two equations.

     \begin{align*}   d x  & = r( \theta cos \theta ) \\   d y  & = r ( - \theta sin \theta ) \\ \end{align*}

So with this information we can construct an integral, perform the necessary substitution and solve.

     \begin{align*}   A & = \frac{\pi (\pi r ) ^2}{2} + \int_{\pi}^0{ \left[   xdy + ydx \right] } \,\mathrm{d} \theta \\    & = \frac{\pi ^ 3  r^2 }{2} + \int_{\pi}^0{ \left[ r(cos \theta + \theta sin \theta) \cdot r( - \theta sin \theta ) +    r(sin \theta - \theta cos \theta ) \cdot r(\theta cos \theta)  \right] } \,\mathrm{d} \theta \\    & = \frac{\pi ^ 3  r^2 }{2} + r^2 \int_{\pi}^0{ \left[ - \theta sin \theta cos \theta - \theta ^ 2 sin ^ 2 \theta  + \theta sin \theta cos \theta - \theta ^ 2 cos ^ 2 \theta  \right] } \,\mathrm{d} \theta \\     & = \frac{\pi ^ 3  r^2 }{2} + r^2 \int_{\pi}^0{ \left[  - \theta ^ 2 ( sin ^ 2 \theta + cos ^ 2 \theta ) \right] } \,\mathrm{d} \theta \\   & = \frac{\pi ^ 3  r^2 }{2} - r^2 \int_{\pi}^0{ \left[ \theta ^2  \right] } \,\mathrm{d} \theta \\   & = \frac{\pi ^ 3  r^2 }{2} - r^2 \left[ \frac{\theta ^3}{3}  \right] _{\pi}^0  \\   & = \frac{\pi ^ 3  r^2 }{2} - r^2 \left[ (0) - (\frac{\pi ^ 3}{3}) \right]   \\  &= \frac{\pi ^ 3  r^2 }{2}  + \frac{r^2 \pi^3}{3} \\ &= \frac{5}{6} \pi^3 r^2 \\ \end{align*}

So finally we have determined the total grazing area of the cow.

In conclusion, as previously discussed, calculus can be incredibly useful to solve a number of problems. In the future we will see just how important calculus is in quantum theory. In my next post I introduce the concept of the paradox behind Gabriel’s horn. This problem will make use of partial differentiation in its application to determine the area of curved surfaces.

Dan Roy, Creator and Founder of Universal Physics.

Category(s): Mathematics
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30 Responses to The Involute of a Circle

  1. thanks for share!

    Tom Laidlaw says:

    I have seen this problem many times, but it always seems like throwing a dart and then drawing the target. Every solution I have seen limits the rope to the half circumference. This makes it nice and neat, but what if I want a longer rope. Is there a way to figure this mathematically with a radius of 10 and a rope of 50. I have done it with a BASIC program, but I would like to know if it can be done mathematically.

    Tom Laidlaw

  2. Your picture has points labeled P, T, and O, but you refer to angles PTR, PTQ, and QRT. What am I missing?

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